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2 years ago

Two 7.0cm×7.0cm metal electrodes are spaced 1.0 mm apart and connected by wires to the terminals of a 9.0 v battery.

Physics

1 answer:

AleksandrR [38]2 years ago

5 0

Answer:

$q=390.285x10^{-12}C$

Explanation:

Those kind of problems of electric physics is about capacitors, so the normal questions are:

What are the charge on each electrode?

Solve this you can get other information required in the problem or write down the other questions you need

$7.0cm*\frac{1m}{100cm}=70x10^{-3}m$

$A=70x10^{-3}m*70x10^{-3}m=4.9x10^{-3}m62$

Capacitance

$C=E_o*\frac{A}{d}$

$E_o=8.85x10^{-12}\frac{C^2}{N*m^2}$

$C=8.85x10^{-12}\frac{C^2}{N*m^2}*\frac{4.9x10^{-3}m^2}{1x10^{-3}m}$

$C=43.365x10^{-12}F$

$C=43.365pF$

The charge is find by the equation

$q=C*V$

$q=43.365pF*9V$

$q=390.285x10^{-12}C$

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